Consider the following axioms on a coloring of the Euclidean
plane $\mathbb R^2$:

A1. Every point of $\mathbb R^2$ is colored $0$ or $1$.A2. Every line contains points of both colors.

Definitions.

A set of points in the plane is called monochromatic if all points have the same color.

A square is a set of $4$ points forming the corners of a square.

An equilateral triangle is a set of $3$ points forming the corners of an equilateral triangle.

A unitary pair is a set of $2$ points at distance 1.

Question.

Does a plane satisfying Axioms A1 and A2 always contain a monochromatic square?
If the answer is yes, then please provide a (minimal) construction (not an existential proof).
If the answer is no, then please provide an impossibility proof.

Hint 1. [added 5/9/2017]

Does a plane satisfying Axioms A1 and A2 always contain a unitary pair?

Answer to Hint 1. Yes.

Pick point X1(0,0) and X2(0,1). Without loss of generality, we can assume that X1(0,0) has color 0. If both are the same color, you are done. If not, point X3(0,2) is either color, so we either have a unitary pair or a pair at distance 2 (not what we want). It seems we are stuck. Originally I could not improve on this result. I had the following insight.

If the (square) grid does not admit a unitary pair, then it must be a checkerboard pattern! However, we want to admit only finite constructions. We are not allowed to construct this checkerboard and use it in a subsequent step! Bummer. But then I realized that we do not have to know this about the grid. The unitary pair is already close at hand. Proceed as follows. If point X3(0,2) is of color 0, construct a point X' at distance 1 from X3(0,2) and at distance 2 from X1(0,0). We can do this by using a compass and intersect 2 circles. Points X1(0,0), X3(0,2) and X' form an isosceles triangle (please, check this).

If X' is of color 0, it forms a unitary pair with X3(0,2).
If X' is of color 1, look at the midpoint X'' between X' and X1(0,0).
If X'' is of color 0, it forms a unitary pair with X1(0,0); if it is of color 1, it forms a unitary pair with X'. Found it!

This picture shows the type of construction I am talking about. Without loss of generality, points 1 and 2 are colored as shown; if they were of opposite colors, turn the plane 180 degrees. If they were equal, it would be the same as a construction starting with 1, 3, 2, etc. (and if point 2 had been red, we would have already succeeded.) Thus, without loss of generality, we can assume that point 3 is red; if it had been green, the construction proceeds mirrored. For all the following points note that if they had the opposite color, we would have already succeeded in finding a monochromatic equilateral triangle. The picture proves that a construction needs no more than 8 points. Can you find a construction using less than 8 points?

Conjecture.

The answer to the question is yes.

I considered that for $n>1$ a square of size $(n-1)*(n-1)$ contains a grid of $n*n$ points, the corners of the unit squares in it. Using only
the points of this grid, it has $O(n^{3})$ squares in it. However, I
suspect that if $n$ is large enough, there is always at least one
monochromatic square in it. Since we can imagine such a grid multiplied,
translated or rotated anywhere in the plane, it would follow that the
plane has an infinite number of monochromatic squares of (approximately)
any size; in a way the plane is dense with monochromatic squares.Still, I have not been able to come up with a construction that shows the existence of at least one square. The reason is that the number of possible colourings equals $2^{n^{2}}$, ignoring symmetries. For $n=2,3,4,5$ I found colourings that do NOT admit a square. $n=6$ turns out too difficult to do by hand. Now consider $n=10$; the number of colorings is $2^{100} \approx 10^{30}$. To disprove the conjecture, it is sufficient to find just ONE colouring that does not admit a square. But to prove the conjecture, I need to find a monochromatic square in ALL colourings. I couldn't see how to simplify this search. Please, help.

If you want to disprove the conjecture, you can colour the plane in any way you want (as long as axioms A1 and A2 hold) and prove for your colouring that it does not have a monochromatic square. If you can do this, then it is clear that no construction of a monochromatic square is possible. Then it will follow that ALL finite grids can be coloured in such a way that it does NOT have a monochromatic square. It would be quite a powerful result. (And therefore not trivial.)

Note.

A2 eq. no line is monochromatic.

I remember this problem as part of an article or or a book I read during the late 80's, but I lost the reference. The original talked about red and green points. I seem to remember that it was a study or research article about the properties of such a coloured plane. I seem to remember that it contained the construction of a monochromatic square. I could be wrong. I am still intrigued after all these years! It seems to be an interesting problem in its own right.

$\begingroup$You seem to be mixing 'planes' and 'grids'. Are we to see this as a 'grid', with each point being a pixel? That would imply you can only have 'lines' that are horizontal, vertical, or diagonal at 45 degree multiples... (and that by your definition of a square, it might not be possible to trace the outline of a square with 4 lines)$\endgroup$
– Tim CouwelierMar 30 '16 at 8:30

$\begingroup$I don't follow. Are all grid coordinates points? What's your definition of a line? Do all points have lines between them? If not, doesn't that mean there might not even be a square at all? if (0,0) is colored "1" does that mean that (0,1), (1,0), (-1,0) and (0,-1) must be colored "0" by A2?$\endgroup$
– Ivo BeckersMar 30 '16 at 9:19

3

$\begingroup$I think Cuc only constructed a grid in the conjecture, but it wasn't required for the actual question. For the full plane, every point (even irrational ones) needs to be coloured, and all lines are infinite length (I think).$\endgroup$
– LacklubMar 30 '16 at 12:49

6 Answers
6

The Hungarian mathematician Tibor Gallai proved the following theorem in the 1930s (but never published his proof):

Theorem: Let $S$ be a finite subset of $\mathbb{Z}^n$.
Then any coloring of $\mathbb{Z}^n$ with finitely many colors contains a monochromatic subset that is homothetic to $S$, that is, a monochromatic subset of the form $x+yS$ with $x,y\in\mathbb{Z}^n$.

A proof was published later in 1943 by Richard Rado in the Proceedings of the London Mathematical Society 48, pp 122-160.
See for instance http://arxiv.org/abs/1304.3154 for more information on this; or google for the terms "Ramsey theory", "Euclidean", "Gallai".

For $n=2$ and $S=\{(0,0),(0,1),(1,0),(1,1)\}$ and for two colors, Gallai's theorem tells us that there indeed always exists a monochromatic square.

$\begingroup$I do not see why you impose condition A2. The statement (existence of a monochromatic square) remains true, even if one drops A2.$\endgroup$
– GamowMar 30 '16 at 13:52

$\begingroup$I added A2, because it asserts that there exist at least two points of different color. It may simplify some proofs.$\endgroup$
– CucMar 30 '16 at 21:10

$\begingroup$And, thanks @Gamow for the reference! I'm certainly going to look at that proof! It's more puzzling now than it ever was. I did get the feel for the "x + y.S". Note that if the theorem holds for the integers in finite dimension, it certainly holds for any grid of different size. I suppose that unless I study the proof, I won't see a construction of the square? I thought it might be a simple case after all. But now that any S can be found, I am totally intrigued!$\endgroup$
– CucMar 31 '16 at 7:37

1

$\begingroup$I did notice that it is harder to find a square that is just a translation or mulitplication of the unit square, then it is to find a square of any orientation. The theorem is on the one hand more powerful, but on the other hand non-constructive, and does not answer my question. See my added answer (I hope it helps the search).$\endgroup$
– CucMar 31 '16 at 19:19

This took me soooooo long. If you read it, I commend you. If you find fault in it, then I will delete this answer and never look at this problem again.

~~~~

We will prove that if a row containing 5 in a row exists, it must contain a monochromatic square. Then we will iteratively prove for 4, 3, 2, and 1. We only need the grid, not the real numbers. And we don't use $A2$.

Also, it doesn't rely on other theorems, which, while make other proofs elegant, also require a level of sophistication that not all puzzlers may have.

It is, unfortunately, just long.

Instead of colours, I use $0$ and $1$.

Five in a row

Lets assume there is a run of five in a row.

00000

There are a few options for how we create the row below. We cannot have two consecutive 0s, so our options are:

Notice that option 1 has no 0s, options 2-4 have 1 0, options 5-8 have 2 0s, and option 9 has 3 0s.

Case 1

Lets try to add a row on top. You can either start with a 0 or a 1.

Case 1.1

Start with a 0. The next must be a 1 to avoid making a 2x2 square. That means the 4th must be a 0 to avoid a 3x3 square of 1s. Since we can't have two consecutive 0s, the 3rd and 5th must also be 1s.

01101 --1-1
00000 -----
11111 --1-1

But this makes a 3x3 monochrome square.

Case 1.2

So instead, we start with a 1 on top. The 3rd must be a 0 and the 2nd and 4th must be 1s.

1101- -1-1-
00000 -----
11111 -1-1-

Again, we've made a monochrome 3x3 square.

Case 2

If you add a 1, you will end up with the same situation as Case 1.2.

So, we must add a 0. The next must be a 1 meaning 4th must be 0, so the 3rd and 5th are 1s.

01101
00000
11110

Now add another row on the bottom.

If we add a 0, the 3rd column must be a 1, which means the 2nd and 4th must be 0s. But this makes a square, so we must instead add a 1 on the bottom. Thus, the 2nd must be a 0, and the 4th must be a 1, so the 3rd must be 0.

Case 4

Notice a 0 in the 3rd spot on top would create a diamond. So we need a 1 there. You cannot have 1s on either side, so at least one of those must be a 0. Possibly both. The options are:

-010- -011- -110-
00000 00000 00000
11011 11011 11011

The last two are symmetric, so we can ignore the last. Notice that both remaining options require a 1 in the first spot. The first also requires a 1 in the last spot.

If you add a 0 in the 4th spot of a new row on top, you will create a tilted monochromatic square. So this must be a 1 for both. This means a 0 is required in the 1st position for both to avoid a 4x4 square, and 1 in the 3rd to avoid a 3x3 square.

The first option is now trivially eliminated now leaving only the 2nd. We will try to add another row on top.

If you add a 1 on top in the 1st position, then the 3rd must be a 0. This makes a tilted square, so it must be a 0. Thus the 4th must be a 1 and the 3rd again must be a 0 making the same tilted square.

Case 5.3

If the 2nd position on top is a 0, then there is a tilted square. So it must be a 1. Thus, the 3rd must be a 0, making the 1st and 5th 1s. The 4th then must be a 0. This means the 2nd row can be completed with a 0.

Lets add another row on top. If we start with a 0, then the 3rd and 4th positions must be a 1s which make a small square. Thus, the row on top must start with a 1, making the 5th spot a 0. Then the 3rd must be a 1, so the 2nd and 4th are 0s.

Case 7

Lets add a row above. If you start with a 1, then the 3rd spot must be a 0 with 1s on either side. If we complete this row with a 1 then this is Case 4. If we complete it with a 0, then it is Case 5. Both of which we've already proven.

1101-
00000
10110

So, we need to start with a 0. Thus, the 2nd spot must be a 1. If we complete the row with 011 or 111, then we will have Case 5 or Case 2. So we must use something else. Options are 010, 101, and 110.

01010 01101 01110
00000 00000 00000
10110 10110 10110

Case 7.1

First note that a 1 is required in the 6th position of the 2nd row to avoid a diamond. Thus, a 1 in the 5th position of the row below would make a tilted square, so it must be a 0. Thus, the 3rd position must be a 1 and the 4th a 0, which also is a tilted square.

Case 8

If the 3rd spot is a 0, then there must be 1s on either side. Therefore, it is a 1.

-101- -1-1- --1--
00000 ----- --> 00000
01110 -1-1- 01110

If there are 0s on either side, then 1s are required on the ends, which makes this Case 6. We've already seen they can't be 1s on either side, so we need one of each. Since it is symmetric, WLOG we can put the 1 in the 2nd spot and a 0 in the 4th. That means a 1 in the 5th.

-1101
00000
01110

Completing this row with a 1 is Case 3 and with a 0 is Case 7.

Case 9

A 0 in the middle position requires 1s everywhere else. This is exactly the Case 4.

11011 --1--
00000 --> 00000
01010 01010

If you put 1s on either side, you create a square of 1s. 0s on either side means you need to finish with 1s, which is Case 6. Thus, we need one of each. Since it is symmetric, WLOG we will pick the 2nd spot to have a 1 and the 3rd a 0. Thus the 5th must be a 1.

-1101
00000
01010

Completing with a 1 is Case 3, and completing with a 0 is Case 7.

Four in a row

Now that we have proven that there is no run of 5, we know at most there is a run of 4 bookended with 1s as follows.

Case 1

Try to add a row on top.

If we add a 1 in the 2nd position, then you need a 0 in the 4th and 1s on either side. This makes a square, so we know that we must start with a 0 in the 2nd position. Then the 3rd must be a 1 and the 5th a 0, making the 4th a 1.

-1101- --1-1- -0110-
100001 ------ --> 100001
-1111- --1-1- -1111-

If we add a 1 in either of the ends, we get a tilted square. But if they are 0s, then the middle two must be 1s.

Going to the row below, if there is a 1 in the 5th spot, we get a tilted square, so it must be a 0. Thus, a 1 is needed in the 3rd spot, and a 0 in the 2nd, and a 1 in the 4th. Also, a 1 is needed in the 4th row 6th spot to prevent a 0 diamond.

If there is a 0 in the 2nd column of the 2nd row, then a 1 is needed in the 3rd column. This means a 0 in the 5th. In the row above, a 0 is needed in the 3rd column. But this makes a tilted square, so we know the 2nd column of the 2nd row must be a 1.

The 3rd spot of the 2nd row cannot be a 1 because we'd need a 0 above it, and a 1 to avoid the tilted square we just saw, which means a 0 in the top row 5th column, which also makes a square. Thus, we need a 0 in the 3rd spot of the 2nd row. Then we need a 1 in the 1st spot of the 2nd row to avoid a diamond, and a 0 in the bottom left corner to avoid a square.

Putting a 0 in the 3rd spot in the top row results in a tilted square, so it must be a 1 with a 0 below it, and a 1 to the left of that. A 1 is needed in the 1st spot of the 2nd row to avoid a diamond.

Notice at this point that there is a similarity highlighted. Because of the symmetry and the fact that all the placement so far has been forced, we can extend the same pattern below. Also note we need 0s in the top right corners to avoid a 4x4 square of 1s. By symmetry, it is also needed in the bottom left.

Notice the highlighted section is simply the inverse of Case 3 on its side.

Case 5

One of the 3rd or 4th must be a 1 (they can't both be 0s). WLOG we will say the 3rd is. Putting a 1 in the 5th row above this results in a tilted square, so it must be a 0. Thus, there is a 1 in the 3rd column.

If you put a 1 in the 4th column of the 2nd row, then a 0 must go above it, and a 1 must be in the 2nd column on top. This makes a tilted square, so we know a 0 must go in the 4th column of the 2nd row. A 1 must go in the 5th. If we put a 1 in the 2nd column of the 2nd row, then we have Case 3 again, so it must be a 0. We can then also put a 1 in the 6th column of the 2nd row to prevent a diamond.

Three in a row

We know any solution cannot have more than three in a row, so it as at most the following:

10001

The possibilities for the row below are:

1. 2. 3. 4.
10001 10001 10001 10001
-111- -110- -101- -010-

Case 1

We know that there are no runs of 4, so the row of ones must have 0s on either side.

For the top row, a 0 in the 3rd position would necessitate 1s on either side making a square. Thus, it needs to be a 1. Either the 2nd or 4th must be a 0 (they can't both be 1s) so WLOG we will assume the 2nd is a 0. Similarly, in the row below, a 1 in the 3rd position makes a 0 square. So it must be a 0.

Now notice the section that has been highlighted. If we put a 1 in the top middle, we will have created Case 1 on the diagonal. Thus it must be a 0.

If you put a 0 in the 5th spot of the 2nd row, we have a tilted square, so this must be a 1. This gives 3 1s in a row, so we must end it in a 0. Also, we need a 0 in the 4th row 5th spot to avoid a square of 1s.

Case 2.2

Thus, in the row above, we must have a 1 in the 2nd spot.

-1---
10001
-110-

If we put a 0 in the 4th spot, then you need a 1 in the 3rd and we have 3 0s in a row, so they must be ended in 1s. But this makes a tilted square, so a 1 is required in the 4th spot. We know the middle must be a 0 (to avoid Case 1).

-110- --1-- -101-
10001 ----1 --> 10001
-110- -1--- -110-
---1- ---1-

The row above cannot have a 1 in the 4th spot due to a tilted square. So it must be a 0. Thus, a 1 is required in the 2nd spot.

Case 4

We know that we can't put 111, 110, or 101 on the top since those scenarios have already been disproven. The only remaining option is 010.

-010- -0-0-
10001 -----
-010- -0-0-

Two in a row

Lets look at boards where the longest length is 2 in a row.

1001

There are two options for the row below;

1. 2.
1001 1001
-11- -01-

Case 1

Adding a single 1 to the 2nd row makes it similar to Case 2 of the "Three in a Row". So we know we must book-end it with 0s.

1001
0110

If we add a 0 in the 2nd spot on top, then we need a 1 in the 3rd. But notice how this is again the same as Case 2 of "Three in a Row" on the diagonal. Thus, we cannot add a 0, but instead it must be a 1. By symmetry, so must the 3rd spot. Again, we must bookend the 1s with 0s.

-01- -0-- 0110
1001 1-0- --> 1001
0110 -1-0 0110

To avoid a diamond of 0s, we need 1s on the ends of the 2nd row.

-0110-
110011
-0110-

You can see that the original pattern is repeating itself above and below. So it must be extended.

Case 2

In order to avoid 3 0s in a row, we need a 1s in the 2nd column of the rows above and below. Similarly, we need 1s on the diagonal to avoid more than 2 0s in a row on that diagonal.

-1-1
1001
-01-
11--

To avoid more than 2 1s in a row, we can add some 0s.

-1-1 ----
1001 --0-
-010 -0-0
110- --0-

One in a Row

Trivially, we can see that the pattern doesn't work.

101 1-1
010 ---
101 1-1

Conclusion

So we have shown that there are no solutions containing 5 in a row, and all solutions with 4 in a row lead to contradictions or 5 in a row. Similarly for 3 in a row and 2 in a row. Lastly, the trivial pattern with 1 in a row is obviously not a solution either.

Therefore, there are no colourings of the grid (and thus the plane) that do not permit a monochromatic square.

$\begingroup$Woh I hope for you this is the good answer !$\endgroup$
– FabichApr 11 '16 at 21:31

$\begingroup$@Trenin, KUDOS! You've successfully shown that there always exist monochromatic squares, and gave a construction! You even corrected my 5x5 square. I had a similar idea I was working on (you beat me), building a decision tree, just like you did. But your idea of considering monochromatic lines is cool, so you could use it in subsequent cases. -- 2 mini mistakes (I read all) without impacting the correctness of your method. "Four in a row" case 4, you can't use symmetry for the last two corners. "Three in a row" case 2.2(.2) can be simplified; the 1 on the second diagonal is forced. Good job!$\endgroup$
– CucApr 11 '16 at 21:40

$\begingroup$@Cuc Are you talking about the placement of 1s in or 0s in the corners for "4 in a row" Case 4? Because if it is for 0s, it is to avoid a 4x4 square of 1s, and if is it for 1s, it is to avoid a 5x5 square of 0s. Since the 6x5 grid is still symmetric, we you can see it is needed in both corners. I will look at the 3x3 case.$\endgroup$
– TreninApr 12 '16 at 12:02

$\begingroup$@Cuc I make Case 4 more clear and took your advice on simplifying Case 2.2(.2).$\endgroup$
– TreninApr 12 '16 at 13:24

$\begingroup$@Trenin. I have been thinking some more on your proof. It too is an existence proof, and I was too fast to call it a construction. However, given the constructive nature of all the sub-cases, and the fact that we know from the theorem that there must be 5 points in a row, perhaps the construction of the 5 in a row is easier than that of a square? I have been trying and got as far as 4 in a row. -- Another idea is to try and construct 5 in a row while trying to prevent a square. We expect that that won't work (because of your solution), so this might give us the construction of a square. Ideas?$\endgroup$
– CucApr 13 '16 at 2:41

Edit: The following is incorrect. I misread the question, and thought that a square included all of the points inside the square. However, only the corners are of interest. Thanks to Mike Earnest for pointing this out.

I think that:

Not all planes have a monochromatic square

Because:

Consider the plane where, for any pair (x,y) it is coloured 0 if $\sqrt{x^2 + y^2}$ is rational. Otherwise, it is coloured 1. This creates "infinitely dense" rings of both colours centred around the origin.

In this plane:

It is clear to see that every point is coloured, so it satisfies A1. Any line of any finite length goes through different radii, because lines of constant radii are curved. If a line goes through different radii, then some radii must be irrational and some must be rational, so the line satisfies A2.

Construction of squares:

It is impossible to construct a monochromatic square in this plane. Any square must have a line segment of finite length, which means it must go through different radii. Going through different radii implies different colours (again, because some radii will be irrational and some will be rational). This means that any square (or indeed any polygon) must not be monochromatic.

Note: I have assumed that you are talking about the plane $R^2$, not only the integer grid $Z^2$.

$\begingroup$When Cuc refers to a "square," what he means is "the four corners of a square," so proving that the line segment is bichromatic doesn't prove that the "square" is$\endgroup$
– Mike EarnestMar 30 '16 at 13:32

$\begingroup$@MikeEarnest Well darn. I'm an idiot. Here I was trying to not have any square regions... It's pretty easy to see that any square centred at the origin will be monochromatic then.$\endgroup$
– LacklubMar 30 '16 at 13:36

$\begingroup$Hi @Lacklub, I appreciate your efforts. But a square centered around the origin is not monochromatic for any coloring. Consider the following pattern 010\ 110\ 001\ It obviously does not contain a monochromatic square.$\endgroup$
– CucMar 31 '16 at 7:39

$\begingroup$@Cuc I meant that in the colouring I proposed, any square centred around the origin is monochromatic. ie. I messed up.$\endgroup$
– LacklubMar 31 '16 at 12:17

It contains Gallai's theorem (Theorem 3.6) and also a proof that to construct a square, we need an nxn grid with 13 <= n < min{VW(8), 5.2^(2^40)} and where VW(8) is the so-called classical Van der Waerden number for n points. The best bounds for VW(8) are 11495 <= VW(8) <= 2^(2^(2^(2^(2^17)))).

So, it seemed that nobody had found a construction of the square yet! I was going to reproduce the graph that showed we need a grid of at least size 13, since there exists a grid of size 12 that does not contain a monochromatic square. But then I noticed that their definition of square seems to only include squares that are positioned orthogonal to the grid. Bummer!

Their grid does contain monochromatic squares, such of the form {A, A+(2,2), A+(4,0), A+(2,-2)}.

I guess that my question isn't answered yet. But here's the proof--in the form of a coloring--that we need a grid of at least size 6 to be able to construct a square for all colorings. None of its 50 squares is monochromatic. For n=6 a grid of 6x6 contains 105 squares, etc.

(thanks to @Trenin)

So, to clarify my idea, if for ALL colorings of a large enough grid there is ALWAYS a monochromatic square, then we can use that pattern to find a monochromatic square in the plane by overlaying the grid ANYWHERE in the plane. That is de facto a construction of a monochromatic square. On the other hand, if we find ONE coloring of a grid that does not have ANY monochromatic square, then we can not use that size of grid to find a monochromatic square . . .

$\begingroup$Your example does have a monochromatic square. The points $(3,2), (4,4), (2,5), (1,3)$ are all red and form a square. If you choose to number the rows starting at 0 instead, then those points are $(2,1), (3,3), (1,4), (0,2)$.$\endgroup$
– TreninApr 7 '16 at 14:01

1

$\begingroup$Here is an example of a 5x5 square with no monochromatic squares: 01101,00000,10110,11100,00111. Sorry for the formatting, but if you draw it out it might help.$\endgroup$
– TreninApr 7 '16 at 14:06

$\begingroup$The best I've been able to do is a 6x5 square. This is seen in "4 in a row" case 4. It is symmetric and adding either a 0 or 1 in the 3rd spot below creates tilted squares (4x4 for a 0 and 6x6 for the 1) so it can't be expanded any further into a 6x6. I would guess that a 6x6 square is impossible.$\endgroup$
– TreninApr 12 '16 at 13:50

$\begingroup$I did a computer search. There are 6x6 colourings with no monochromatic squares. If the program is correct, there are only 56 of them, but by rotation/reflection/inversion that reduces to only 5 essentially different ones. Here's one: 011101, 110000, 011010, 010110, 000011, 101110. All the ways of merging two of these 6x6 colourings together to form a 7x6 rectangle colouring results in a colouring with monochromatic squares in it, so there are no larger colourings without monochromatic squares.$\endgroup$
– Jaap ScherphuisMay 10 '17 at 12:08

$\begingroup$@JaapScherphuis Nice work! I like how your example is symmetric.$\endgroup$
– TreninMay 10 '17 at 12:25

If you consider a 'square' to be a filled square or even only the outline (= lines only) of a square, then No.

Extend a chess board to infinity -- there will never be a square of same-colored elements.

Proof:

Assume your points have integer coordinates $((1,1), (1,0), (12345,6342)$, etc.). Then you get a chess board pattern with the following coloring rule: $x+y$ even => color $0$; $x+y$ odd => color $1$.This means that, no matter in which direction you move, the next point will always have the opposite color (since you are only changing $1$ coordinate, and so going from odd to even or vice versa). But to create a non-trivial square (either outline or filled) you need to have at least two neighboring points of the same color.
If you don't have integer coordinates but real coordinates, Lacklub has given you the answer of constructing such a plane.

If you consider a square to be a set of $4$ points of which always two share either an $x$ or a $y$ coordinate (= corners of a rectangle), then YES as long as $n>4$

No for $n=4$:

$1100$
$1010$
$0101$
$0011$

EDIT: The following lemma and proof are for RECTANGLES, not for SQUARES!!!

Lemma:

if you have an $n*n$ plane, you can sort rows and columns as you like without the square-or-not-square property being affected. (obvious)

Proof for YES:

Assuming you have $n\ge4$ then you will have a row where one color appears $3$ or more times. Let this color be $0$. Since you can sort the columns, you can assume that the $0s$ are next to each other:
$000$
To prevent a square from forming, all other rows may only have one $0$ at the most (if you have $2$ or $3$ zeroes, they would form a square with the first row).
$000$
$110$
$101$
$011$
Now the next row needs to have two $1s$.
However, there are only $3$ distinct possibilities of distributing two indistinguishable items in three slots.
So with the next row $(n=5)$ you will inevitably repeat one of the patterns, leading to a square of $1s$.
Ergo, with $n\ge5$ you will have at least one such 'square'.

$\begingroup$That is a rectangle. I think a square must have all sides the same length.$\endgroup$
– TreninMar 30 '16 at 14:45

$\begingroup$Hi @subrunner, I consider a square to be ANY set of 4 points that form a square. Even in a grid, there are many. Take any point A on the grid and a vector (x,y) with x, y integers. Now look at the points {A, A+(x,y), A+(x,y)+(-y,x), A+(x,y)+(-y,x)+(-x,-y)}. They are the corners of a square. Etc.$\endgroup$
– CucMar 31 '16 at 7:44

As an answer to the hint, a construction of an equilateral triangle can be done with no more than 7 points. The following diagrams share points 1-3 of the construction, and the construction follows Fig. 1 if point 4 is green, and follows Fig. 2 if point 4 is red. Note that the used colorings can be assumed without loss of generality. Further, if point 2 is red, consider it point 3 and rotate the diagram to proceed the construction with points 2, 4, etc. If points 1, 2 and 3 are of the same color or any point beyond 4 is of the other color, a monochromatic equilateral triangle has already been found using less than 7 points and the construction stops. The construction needs not more than 7 points. Are there any constructions of monochromatic equilateral triangles using less points?